Three Dimensional Geometry Question 9
Question 9 - 2024 (27 Jan Shift 2)
Let the position vectors of the vertices $A, B$ and $C$ of a triangle be $2 \hat{i}+2 \hat{j}+\hat{k},{ }^{\prime}+2 \hat{j}+2 \hat{k}$ and $2 \hat{i}+\hat{j}+2 \hat{k}$ respectively. Let $l_{1}, l_{2}$ and $l_{3}$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ respectively, then $l_{1}^{2}+l_{2}^{2}+l_{3}^{2}$ equals:
(1) $\frac{1}{5}$
(2) $\frac{1}{2}$
(3) $\frac{1}{4}$
(4) $\frac{1}{3}$
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Answer (2)
Solution
$\triangle \mathrm{ABC}$ is equilateral
Orthocentre and centroid will be same
$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$
Mid-point of AB is D $\left(\frac{3}{2}, 2, \frac{3}{2}\right)$
$\therefore \ell_{1}=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}$
$\ell_{1}=\sqrt{\frac{1}{6}}=\ell_{2}=\ell_{3}$
$\therefore \ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2}=\frac{1}{2}$
